Annales Academiæ Scientiarum Fennicæ
Mathematica
Volumen 34, 2009, 583–605
INITIAL-BOUNDARY VALUE PROBLEMS
FOR PARABOLIC EQUATIONS
Magnus Fontes
Lund University, Centre for Mathematical Sciences
Box 118, SE-221 00 Lund, Sweden; [email protected]
Abstract. We prove new existence and uniqueness results for weak solutions to non-homogeneous initial-boundary value problems for parabolic equations modeled on the evolution of the
p-Laplacian.
1. Introduction
In this paper we prove new existence and uniqueness results for weak solutions
to non-homogeneous initial-boundary value problems for parabolic equations of the
form
∂u
(1.1a)
− ∇x · A(x, t, ∇x u) = f in D 0 (Q+ ),
∂t
(1.1b)
u = g on (Ω × {0}) ∪ (∂Ω × R+ ).
Here Ω is an open and bounded set in Rn and Q+ = Ω × R+ . Precise structural conditions for A(·, ·, ·) are given in Section 4, but the model is the following p-parabolic
equation
∂u
(1.2a)
− ∇x · (|∇x u|p−2 ∇x u) = f in D 0 (Q+ ),
∂t
(1.2b)
u = g on (Ω × {0}) ∪ (∂Ω × R+ ),
with 1 < p < ∞.
The boundary data is prescribed on the whole parabolic boundary, (Ω × {0}) ∪
(∂Ω × R+ ), and we study the problem of finding the “largest possible” classes of
boundary and source data such that (1.1) has a good meaning and is uniquely solvable.
In the case of the elliptic p-laplacian:
(1.3a)
−∇ · (|∇u|p−2 ∇u) = f
(1.3b)
u=g
in D 0 (Ω),
on ∂Ω,
it is well known that W 1,p (Ω) is a kind of golden mean. It has the useful property
that:
Given g ∈ W 1,p (Ω), there exists a unique solution u ∈ W 1,p (Ω) to the p-laplace
equation (1.3) such that u − g belongs to the closure of D(Ω) in the W 1,p (Ω)-norm
topology. Furthermore the source data (f in (1.3)) can then be taken as sums of first
order derivatives of Lp/(p−1) (Ω)-functions.
2000 Mathematics Subject Classification: Primary 35K60, 35K55.
Key words: Parabolic equations, nonhomogeneous boundary conditions, nonhomogeneous initial data, Hilbert transform, anisotropic Sobolev spaces, evolution of p-Laplacian.
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Magnus Fontes
In this paper we construct an analogous optimal solution-space for equations of
the type (1.1).
We point out that our results are new even in the linear case. In the linear case,
where p = 2 and we denote W s,2 by H s , it is well known (see, e.g., [5] Vol. II) that the
parabolic solution and lateral boundary value spaces, replacing the “elliptic spaces”
H s (Ω) and H s−1/2 (∂Ω), are H s,s/2 (Ω × R+ ) and H s−1/2,s/2−1/4 (∂Ω × R+ ). The initial
data on Ω × {0} should then belong to H s−1 (Ω) and the natural source data space
is H s−2,s/2−1 (Ω × R+ ). With additional compatibility conditions for the coupling of
the data in the “corners” of the space-time cylinder we then have unique solvability
for the linear case when s > 1 (see [5], Vol. II). When s = 1, the golden mean in the
elliptic case, several difficulties arise in the parabolic case. One obvious difficulty is
of course that we are in the borderline Sobolev imbedding case in the time direction
(half-a-time derivative in L2 (R+ , L2 (Ω))), and are thus for instance unable to define
traces on Ω × {0}.
In Theorem 4.10 we give optimal results in the linear limiting case (s = 1), and a
complete description of the space of solutions (compare with the non-optimal results
in, e.g., [5],[4] and [3]).
We use a similar construction of the solution space (with new technical complications) in the non-linear case when p 6= 2.
Our solution space for a general p, 1 < p < ∞, (see Definition 4.6) is the sum
of a Banach space carrying initial data and another Banach space carrying lateral
boundary data. It is a dense subspace of the space of Lp (Q+ )-functions, having half
order time derivatives in L2 (Q+ ) and first order space derivatives in Lp (Q+ ).
This statement requires some explanation and the appropriate distribution theory, allowing fractional differentiation in the time direction of general Lp -functions in
a space-time half cylinder, is developed. This analytic framework makes it possible to
give a precise meaning to the fractional integration by parts for the time derivatives
that is one of the key tools in our method. We point out that we use two different
half-a-time derivatives (adjoint to each other) and that demanding these different
derivatives to belong to L2 (Q+ ) gives rise to different function spaces. In Section 4
we investigate the relations between these different function spaces and discuss some
of their basic properties. It is for instance non-trivial to show that our function spaces
are well behaved when we cut off (in a smooth way) in time. This is, apart from the
fact that we are in the borderline Sobolev imbedding case in the time direction, due
to the fact that they have non-homogeneous summability and regularity conditions,
and that they are defined as spaces of distributions.
Most of these technical problems arise already for functions defined on the real
line and half-line, and for clarity we have moved most of these arguments to an
auxiliary section (Section 3) dealing with this case.
The main result of this paper is Theorem 4.8 which implies, among other things,
that our solution space X 1,1/2 (Q+ ) really is a true analog of the space W 1,p (Ω) for
the elliptic p-laplacian, in the sense that:
Given g ∈ X 1,1/2 (Q+ ) there exists a unique solution u ∈ X 1,1/2 (Q+ ) to the pparabolic equation (1.1) such that u − g belongs to the closure of D(Q+ ) in the
X 1,1/2 (Q+ )-norm topology. Furthermore the source data (f in (1.1)) can be taken
as sums of first order space derivatives of Lp/(p−1) (Q+ )-functions and half-a-time
derivatives of L2 (Q+ )-functions.
Initial-boundary value problems for parabolic equations
585
For simplicity we shall assume throughout the paper that the boundary of Ω is
smooth, but this assumption is only used to prove that we can regularize functions
near the lateral boundary so that the different spaces of test functions we use are
dense in the corresponding function spaces (see Theorem 4.1).
2. Some analytical background
We will use the fractional calculus presented in [1]. Here we first give a brief
review of the notation and some results. We then extend the calculus to space-time
half-cylinders in order to be able to discuss initial-boundary value problems.
The Fourier transform on the Schwartz class S (Rn , C) is defined by
Z
u(x)e−i2πx·ξ dx, u ∈ S (Rn , C).
(2.1)
û(ξ) =
Rn
The inverse will be denoted
(2.2)
Z
u(x)ei2πx·ξ dx,
ǔ(ξ) =
u ∈ S (Rn , C).
Rn
The isotropic fractional Sobolev spaces are defined as follows.
Definition 2.1. For s ∈ R and 1 < p < ∞ let
(2.3)
Hps (Rn , C) = {u ∈ S 0 (Rn , C); ((1 + |2πξ|2 )s/2 û(ξ))∨ ∈ Lp (Rn , C)}.
They are separable and reflexive Banach spaces with the obvious norms. We
will use the following multi-index notation. Let α = (α1 , . . . , αn ) ∈ Rn be an ntuple. We write α > 0 if αj > 0, j = 1, . . . , n; xα = xα1 1 · · · xαnn when x ∈ Rn ;
xα+ = xα1 1 + · · · xαnn + , (where t+ = max(0, t) for t ∈ R, with a similar definition for
xα− ) and Γ(α) = Γ(α1 ) · · · Γ(αn ), where Γ denotes the gamma function. Furthermore
we will sometimes write k for the multi-index (k, . . . , k), the interpretation should be
clear from the context. We now define the classical Riemann–Liouville convolution
operators.
Definition 2.2. For a multi-index α > 0, set
(2.4)
−α
D±
u = χα−1
∗ u,
±
u ∈ S (Rn , C),
where the kernels χα−1
± , are given by
(2.5)
χα−1
= Γ(α)−1 (·)α−1
±
± .
α
to general multi-indices α ∈ Rn in the usual way.
We extend the definition of D±
Definition 2.3. For α ∈ Rn set
(2.6)
α−k
α
u,
u = D k D±
D±
u ∈ S (Rn , C),
where we choose the multi-index k ∈ {0, 1, 2, . . . }n so that k − α > 0.
The definition is independent of the choice of k.
Although it is clear in this setting how the support of a function is affected under
these mappings and also for instance that the operators map real valued functions
to real valued functions, other features become transparent on the Fourier transform
side.
Computing in S 0 (Rn , C), we have for all α ∈ Rn :
(2.7)
α
u = ((0±i2πξ)α û(ξ))∨ ,
D±
u ∈ S (Rn , C).
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Magnus Fontes
We will use the following space of test functions.
Definition 2.4. Let
n
o
(2.8)
F (Rn , C) = u ∈ C ∞ (Rn , C); kukHps (Rn ,C) < ∞, s ∈ R, 1 < p < ∞ .
F (Rn , C) becomes a Fréchet space with the topology generated by, for instance,
the following family of semi-norms k · kHps (Rn ,C) , s ∈ {0, 1, 2, . . . }, p = 1 + 2k , k ∈ Z.
We have the following dense continuous imbeddings,
(2.9)
D(Rn , C) ,→ S (Rn , C) ,→ F (Rn , C) ,→ E (Rn , C).
An example of a function that belongs to F (R, C) but does not belong to
S (R, C) is x 7→ 1/(1 + x2 ).
For α ≥ 0 we now define the fractional derivatives
α
D±
u = ((0±i2πξ)α û)∨ ,
(2.10)
u ∈ F (Rn , C).
α
α
The operators D+
and D−
are adjoint to each other and they are connected through
the operator
α
(2.11)
H =
n
Y
(cos(παk )Id + sin(παk )Hk ),
k=1
where Id is the identity operator and Hk is the Hilbert transform with respect to the
kth variable, i.e.,
Z
u(t − sek )
−1
(2.12)
Hk u(t) = π lim
ds, u ∈ F (Rn , C),
ε→+0 |s|≥ε
s
where ek is the usual canonical kth basis vector in Rn . We have the following lemma.
α
Lemma 2.1. For α ≥ 0, D±
are continuous linear operators on F (Rn , C). For
α ∈ Rn , H α is an isomorphism on F (Rn , C). For α, β ≥ 0 we have
(2.13)
α+β
α β
,
D±
D± = D±
(2.14)
α α
α
D+
H = D−
.
Furthermore, all these operators commute on F (Rn , C).
We note that for α ≥ 0
Z
Z
α
(2.15)
D+ uΦ dx =
Rn
and for α ∈ Rn
Z
Rn
α
uD−
Φ dx,
u, Φ ∈ F (Rn , C),
Z
α
(2.16)
uH −α Φ dx,
H uΦ dx =
Rn
u, Φ ∈ F (Rn , C).
Rn
Now let F 0 (Rn , C) denote the space of continuous linear functionals on F (Rn , C),
endowed with the weak∗ topology.
α
and H α to F 0 (Rn , C)
Inspired by (2.15) and (2.16), we extend the definition of D±
by duality in the obvious way.
Definition 2.5. For u ∈ F 0 (Rn , C) and α ≥ 0 let
(2.17)
α
α
Φi,
u, Φi := hu, D∓
hD±
Φ ∈ F (Rn , C),
Initial-boundary value problems for parabolic equations
587
and for α ∈ Rn let
(2.18)
hH α u, Φi := hu, H −α Φi,
Φ ∈ F (Rn , C).
The counterpart of Lemma 2.1 is valid for F 0 (Rn , C).
α
Lemma 2.2. For α ≥ 0, D±
are continuous linear operators on F 0 (Rn , C). For
n
α
α ∈ R , H is an isomorphism on F 0 (Rn , C). For α, β ≥ 0 we have
(2.19)
α+β
α β
D±
D± = D±
,
(2.20)
α α
α
D+
H = D−
.
Furthermore, all these operators commute on F 0 (Rn , C).
α
We recall that D±
and H α all take real-valued functions (distributions) to realvalued functions (distributions), and from now on all functions and distributions will
be real valued. We will denote the subspaces of real-valued functions and distributions
simply by F (Rn ) and F 0 (Rn ).
In [1] we studied parabolic operators on a space-time cylinder Q = Ω × R, where
Ω was a connected and open set in Rn . We then introduced the following space of
test functions.
Definition 2.6. Let F0,· (Q) denote the subspace of F (Rn × R) functions with
support in K × R for some compact subset K ⊂ Ω.
We put a pseudo-topology on F0,· (Q) by specifying what sequential convergence
means. We say that Φi → 0 in F0,· (Q) if and only if the supports of all Φi ’s are
contained in a fixed set K×R, where K ⊂ Ω is a compact subset, and kDα Φi kLP (Q) →
0 as i → ∞ for all multi-indices α ∈ Zn+1
and 1 < p < ∞.
+
The corresponding space of distributions is then defined as follows.
Definition 2.7. If u is a linear functional on F0,· (Q), then u is in F 0 ·,· (Q) if
and only if for every compact set K ⊂ Ω, there exist constants C, p1 , . . . , pN with
1 < pi < ∞, i = 1, . . . , N , and multi-indices α1 , . . . , αN with αi ∈ Zn+1
+ , i = 1, . . . , N ,
such that
(2.21)
|hu, Φi| ≤ C
N
X
kDαi ΦkLpi (Q)
i=1
for all Φ ∈ F0,· (Q) with support in K × R.
The motivation for these spaces is that they are invariant under fractional differentiation and Hilbert-transformation in the time variable, and ordinary differentiation
in the space variables. In the given topologies, these operations are continuous.
For initial-boundary value problems, the parabolic operators will by defined on a
space-time half-cylinder Q+ = Ω × R+ , and we shall then need the following natural
spaces of test functions defined on Q+ .
Remark. We shall use the same constructions on the real line and half-line,
which can be thought of as the case Ω = {0} if we identify {0} × R with R and
{0} × R+ with R+ .
Definition 2.8. Let F0,· (Q+ ) denote the space of those functions defined on Q+
that can be extended to all of Q as elements in F0,· (Q).
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Magnus Fontes
Furthermore, let F0,0 (Q+ ) denote the space of those functions defined on Q+
that can be extended by zero to all of Q as elements in F0,· (Q).
(A zero in the first position of course corresponds to zero boundary data on the
lateral boundary and a zero in the second position corresponds to zero initial data.)
By using the construction in [6] of a (total) extension operator, we see that
F0,· (Q+ ) can be identified with the space of all smooth functions Φ, defined on Q+ ,
with support in K × R+ for some compact subset K ⊂ Ω (i.e., they are zero on
the complement, with respect to Q+ , of K × R+ ), with kDα ΦkLP (Q+ ) < ∞ for all
multi-indices α ∈ Zn+1
and 1 < p < ∞.
+
Thus, we can put an intrinsic pseudo-topology on F0,· (Q+ ) by defining that
Φi → 0 in F0,· (Q+ ) if and only if the supports of all Φi are contained in a fixed set
K × R+ , where K ⊂ Ω is a compact subset, and kDα Φi kLP (Q+ ) → 0 as i → ∞ for
all multi-indices α ∈ Zn+1
and 1 < p < ∞. Then F0,0 (Q+ ) is a closed subspace of
+
F0,· (Q+ ) with the induced topology.
We also note that D(Q+ ) is densely continuously imbedded in F0,0 (Q+ ).
Connected with these spaces of test functions are the following spaces of distributions.
Definition 2.9. If u is a linear functional on F0,· (Q+ ), then u is in F 0 ·,0 (Q+ )
if and only if for every compact set K ⊂ Ω, there exist constants C, p1 , . . . , pN with
1 < pi < ∞, i = 1, . . . , N , and multi-indices α1 , . . . , αN with αi ∈ Zn+1
+ , i = 1, . . . , N ,
such that
(2.22)
|hu, Φi| ≤ C
N
X
kDαi ΦkLpi
i=1
for all Φ ∈ F0,· (Q+ ) with support in K × R+ .
Furthermore, if u is a linear functional on F0,0 (Q+ ), then u is in F 0 ·,· (Q+ ) if
and only if for every compact set K ⊂ Ω, there exist constants C, p1 , . . . , pN with
1 < pi < ∞, i = 1, . . . , N , and multi-indices α1 , . . . , αN with αi ∈ Zn+1
+ , i = 1, . . . , N ,
such that
(2.23)
|hu, Φi| ≤ C
N
X
kDαi ΦkLpi (Q+ )
i=1
for all Φ ∈ F0,0 (Q+ ) with support in K × R+ .
The importance of these spaces comes from the fact that, for a real-valued α ≥ 0,
the operations
∂+α
(0,...,0,α)
(2.24)
:= D+
: F0,0 (Q+ ) −→ F0,0 (Q+ )
α
∂t
∂−α
(0,...,0,α)
(2.25)
:= D−
: F0,· (Q+ ) −→ F0,· (Q+ )
α
∂t
are continuous. Ordinary differentiations with respect to the space variables are
clearly also continuous operations on these spaces. We shall also use that the Hilberttransform in the time variable
(2.26)
h := H (0,...,0,1/2) : F0,0 (Q+ ) −→ F0,· (Q+ ),
is a continuous operator.
Initial-boundary value problems for parabolic equations
589
Extending these operators by duality in the obvious way we get that
∂+α
(2.27)
: F 0 ·,0 (Q+ ) −→ F 0 ·,0 (Q+ ),
∂tα
∂−α
(2.28)
: F 0 ·,· (Q+ ) −→ F 0 ·,· (Q+ ),
α
∂t
(2.29)
h : F 0 ·,0 (Q+ ) −→ F 0 ·,· (Q+ ),
and taking ordinary derivatives in the space variables, are continuous operations.
Using the total extension operator from [6], one can show that we can identify
0
F ·,0 (Q+ ) with the space of F 0 ·,· (Q)-distributions that are zero on Ω × (−∞, 0).
Since D(Q+ ) is densely continuously imbedded in F0,0 (Q+ ), we get that F 0 ·,· (Q+ )
is a continuously imbedded subspace of D 0 (Q+ ).
We remark that the space F 0 ·,0 (Q+ ) contains elements supported on Ω × {0}. In
fact,
(2.30)
F 0 ·,· (Q+ ) ' F 0 ·,0 (Q+ )/F ◦ 0,0 (Q+ ),
where F ◦ 0,0 (Q+ ) = {ξ ∈ F 0 ·,0 (Q+ ); hξ, Φi = 0, Φ ∈ F0,0 (Q+ )}.
Finally, since F0,0 (Q+ ) is densely continuously imbedded in Lp (Q+ ) when 1 <
p < ∞, clearly Lp (Q+ ) is continuously imbedded in both F 0 ·,· (Q+ ) and F 0 ·,0 (Q+ )
when 1 < p < ∞. Thus
∂+α
(2.31)
: Lp (Q+ ) −→ F 0 ·,0 (Q+ )
∂tα
∂−α
(2.32)
: Lp (Q+ ) −→ F 0 ·,· (Q+ ),
∂tα
are well-defined continuous operations when 1 < p < ∞.
3. Auxiliary spaces on the real line and half-line
We shall use the following auxiliary spaces defined on R and in the definition
1/2
∂−
∂t1/2
should be understood in the F 0 (R) distribution sense.
Definition 3.1. For 1 < p < ∞, set
(
(3.1)
B 1/2 (R) =
)
1/2
∂
u
u ∈ Lp (R); −1/2 ∈ L2 (R) .
∂t
We equip these spaces with the following norms.
° ∂ 1/2 u °
°
°
+ kukLp (R) .
(3.2)
kukB 1,1/2 (R) := ° −1/2 °
∂t
L2 (R)
Computing in F 0 (R) we see that we can represent these spaces as closed subspaces of the direct sums L2 (R) ⊕ Lp (R), and thus they are reflexive and separable
Banach spaces in the topologies arising from the given norms.
If {ψε } is a regularizing sequence it is clear that
(3.3)
kψε ∗ ukB 1/2 (R) ≤ kukB 1/2 (R) ,
and thus smooth functions are dense in B 1/2 (R).
Due to the definition using distributions and to the inhomogeniety of our summability conditions, it is unfortunately not so easy to cut off in time and in this way
show that F (R) (or D(R)) is dense in B 1/2 (R). Nevertheless this is true.
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Magnus Fontes
Lemma 3.1. The space of testfunctions F (R) is dense in B 1/2 (R).
Proof. The proof is based on a non-linear version of the Riesz representation
theorem.
1/2
We (temporarily) denote the closure of F (R) in B 1/2 (R) by B0 (R), and we
1/2
shall show that B0 (R) = B 1/2 (R).
Set
∂u
(3.4)
T (u) =
+ |u|p−2 u.
∂t
By fractional integration by parts
Z 1/2 1/2
∂+ u ∂− Φ
(3.5)
hT (u), Φi =
+ |u|p−2 uΦ dt ; Φ ∈ F (R),
1/2 ∂t1/2
R ∂t
and Hölder’s inequality, it is clear that
(3.6)
1/2
T : B 1/2 (R) −→ B0 (R)∗
is continuous.
We notice that
(3.7)
1/2
1/2
T : B0 (R) −→ B0 (R)∗
is weakly continuous and monotone (for definitions see [KS] or [1]).
By M. Riesz’ conjugate function theorem, which says that the Hilbert transform
h is bounded from Lp (R) to Lp (R) (recall that 1 < p < ∞), we see that the operators
1/2
H α introduced above are isomorphisms on B0 (R).
Now for any α ∈ (0, 1/2) we have
Z
1/2
1/2
∂+ u ∂+ u
−α
(3.8)
hT (u), H (u)i ≥
sin(πα) 1/2 1/2 + (cos(πα) − sin(πα)C)|u|p dt;
∂t
∂t
R
u ∈ F (R), where C < ∞ is a constant such that
(3.9)
kh(u)kLp (R) ≤ CkukLp (R) .
Choosing α ∈ (0, 1/2) small enough we see that H α ◦ T is coercive. It follows that T
is a bijection (see [1] for this functional-analytic result and similar arguments).
1/2
Thus given u ∈ B 1/2 (R) there exists a unique v ∈ B0 (R) such that T (u) = T (v)
in F 0 (R), i.e.,
∂(u − v)
+ (|u|p−2 u − |v|p−2 v) = 0.
∂t
This shows that the difference of elements with the same image has more regularity in time, namely ∂(u−v)
∈ Lp/(p−1) (R).
∂t
The class of Lp (R) functions with derivatives in Lp/(p−1) (R) is stable under regularization and thus by a continuity argument we see that we can test with χ(u − v),
where χ is a cut off function in time, in equation (3.10). We get that (for a canonical
continuous representative) t 7→ |u − v|(t) is decreasing. Since u − v belongs to Lp (R),
we conclude that u = v. The lemma follows.
¤
(3.10)
We are now in position to prove the following lemma.
Initial-boundary value problems for parabolic equations
591
Lemma 3.2. If u ∈ B 1/2 (R) then
¯
¯
ZZ
Z ¯¯ 1/2 ¯¯2
¯ u(s) − u(t) ¯2
¯ ∂− u ¯
¯
¯
(3.11)
¯ 1/2 ¯ dt.
¯ s − t ¯ ds dt = 2π
¯
R×R
R ¯ ∂t
Proof. Since F (R) is dense in B 1/2 (R) we can compute using the Fourier transform
Z
Z ¯¯ 1/2 ¯¯2
ZZ
1
|1 − ei2πτ s |2
¯ ∂− u ¯
2
2π|τ ||û| dτ =
(3.12)
|û(τ )|2 dτ ds.
¯ 1/2 ¯ dt =
2
¯
2π
s
R
R ¯ ∂t
R×R
Using Parseval’s formula the lemma follows.
¤
We note the following scaling and translation invariance
¯
¯
ZZ
¯ u(a(s − b)) − u(a(t − b)) ¯2
¯ ds dt
¯
¯
¯
s−t
R×R
(3.13)
¯
¯2
ZZ
¯ u(s) − u(t) ¯
¯
¯
=
¯ s − t ¯ ds dt ; a, b ∈ R.
R×R
We also note the following fact.
Lemma 3.3. The space B 1/2 (R) is continuously imbedded in the space of functions with vanishing mean oscillation, V M O(R).
Proof. Let I ⊂ R denote a bounded interval and let uI denote the mean value
of u ∈ B 1/2 (R) over I. Then by Jensen’s inequality
¯
Z
ZZ ¯
¯ u(s) − u(t) ¯2
1
2
¯
¯
(3.14)
¤
|u − uI | dt ≤
¯ s − t ¯ ds dt.
|I| I
I×I
Using the form of the norm in Lemma 3.2, we can now show that we have good
estimates in the B 1/2 (R)-norm for the following cut-off operation.
Lemma 3.4. Let χn be the piecewise affine function that is one on (−n, n),
zero on (−∞, −2n) ∪ (2n, ∞) and affine in between. Let In = (−2n, 2n) and for
u ∈ B 1/2 (R), denote the mean value of u over In by uIn . Then there exists a
constant C such that
¯
¯
ZZ
¯ χn (u − uIn )(s) − χn (u − uI )(t) ¯2
¯
¯ ds dt
¯
¯
s
−
t
R×R
(3.15)
¯
¯2
ZZ
¯ u(s) − u(t) ¯
¯
¯
≤C
¯ s − t ¯ ds dt,
R×R
(3.16)
kχn (u − uIn )kpLp (R) ≤ CkukpLp (R) ;
u ∈ B 1/2 (R).
Furthermore χn (u − uIn ) → u in B 1/2 (R) as n → ∞.
Proof. The boundedness of the cut-off operation in the Lp -norm follows from
Jensen’s inequality. For the L2 -part of the norm an elementary computation gives us
¯
¯
ZZ
¯ χn (u − uIn )(s) − χn (u − uI )(t) ¯2
¯ ds dt
¯
¯
¯
s−t
R×R
(
)
(3.17)
¯
¯2
Z
ZZ
¯
¯
1
u(s)
−
u(t)
¯
¯
≤C
|u − uIn |2 dt +
¯ s − t ¯ ds dt ,
|In | In
R×R
592
Magnus Fontes
and thus (3.15) follows using (3.14). That χn u → u in Lp (R) is clear. If u has
compact support, since p > 1, using Jensen’s inequality, we see that χn uIn → 0 in
Lp (R). Since by Jensen’s inequality χn uIn is uniformly bounded in Lp (R), a density
argument proves that χn (u − uIn ) → u in Lp (R). That χn (u − uIn ) → u for the
L2 -part of the norm follows since by an elementary computation
¯
¯
¯ (1 − χn )(u − uIn )(s) − (1 − χn )(u − uI )(t) ¯2
¯
¯ ds dt
¯
¯
s
−
t
R×R
)
(
¯
¯2
Z
ZZ
¯
¯
u(s)
−
u(t)
1
¯
¯
≤C
|u − uIn |2 dt +
¯ s − t ¯ ds dt .
|In | In
|t|>n
ZZ
(3.18)
(3.19)
The last term clearly tends to zero as n tends to infinity. We only have to prove that
also
Z
1
(3.20)
|u − uIn |2 dt −→ 0
|In | In
as n → ∞. This is true since
Z
ZZ
1
1
2
|u − uIn | dt ≤ 2
|u(s) − u(t)|2 ds dt
|In | In
4n
( In ×IZn Z
¯
¯
¯ u(s) − u(t) ¯2
log2 n
¯
¯
≤C
(3.21)
¯ s − t ¯ ds dt
n2
|s|,|t|≤log n
)
¯
¯
ZZ
¯ u(s) − u(t) ¯2
¯
¯
+
¯ s − t ¯ ds dt ,
|t|≥log n
which clearly tends to zero as n tends to infinity.
¤
Remark. We subtracted the mean value in the argument above in order not
to have to rely on the fact that u ∈ Lp (R) when proving boundedness for the halfderivatives. This is crucial when we later use the same argument on functions defined
in a space-time cylinder. In preparation for this we also note that, by regularizing,
the lemma gives us an explicit sequence of D(R)-functions tending to a given element
in B 1/2 (R).
We now introduce two sets of spaces defined on the real half-line.
1/2
Definition 3.2. Let B0 (R+ ) be the space of functions defined on R+ that can
be extended by zero as elements in B 1/2 (R).
Furthermore, let B 1/2 (R+ ) be the space of functions defined on R+ that can be
extended as elements in B 1/2 (R).
1/2
Remark. The space B0 (R+ ) can of course be identified with the closed subspace of B 1/2 (R) of functions with support in R+ .
1/2
We now give two simple lemmas, giving intrinsic descriptions of B0 (R+ ) and
B 1/2 (R+ ). We omit the proofs, which are straightforward elementary computations
using the form of the norm in Lemma 3.2.
Initial-boundary value problems for parabolic equations
593
1/2
Lemma 3.5. The function space B0 (R+ ) is precisely the set of Lp (R+ )-functions such that the following norm is bounded:
½Z
u2 (t)
kukB 1/2 (R+ ) := kukLp (R+ ) +
dt
0
t
R+
)1/2
(3.22)
¶2
µ
ZZ
u(s) − u(t)
ds dt
.
+
s−t
R+ ×R+
Lemma 3.6. The function space B 1/2 (R+ ) is precisely the set of Lp (R+ )-functions such that the following norm is bounded:
(Z Z
)1/2
µ
¶2
u(s) − u(t)
.
(3.23)
kukB 1/2 (R+ ) := kukLp (R+ ) +
ds dt
s−t
R+ ×R+
Furthermore, a continuous symmetric extension operator from B 1/2 (R+ ) to B 1/2 (R)
is given by ES (u)(t) = u(|t|).
We have the following density results:
Lemma 3.7. The space F (R+ ) is dense in B 1/2 (R+ ) and F0 (R+ ) is dense in
1/2
B0 (R+ ).
Proof. That F (R+ ) is dense in B 1/2 (R+ ) follows immidiately from the fact that
1/2
F (R) is dense in B 1/2 (R). The argument to prove that F0 (R+ ) is dense in B0 (R+ )
1/2
is a little more delicate. Given u ∈ B0 (R+ ), apriori we only know that there exists
a sequence of testfunctions in F (R) approaching u in the B 1/2 (R)-norm.
1/2
Given u ∈ B0 (R+ ) we will show that we can cut-off. Let χn be the piecewise
affine function that is one on (0, n), zero on (2n, ∞) and affine in between. We will
1/2
show that χn u → u in B0 (R+ ). Taking this for granted we can regularize with a
regularizing sequence having support in R+ which gives us the lemma.
That χn u → u in Lp (R+ ) is clear. We now estimate the L2 -part of the norm. An
elementary computation gives us
Z
ZZ
((1 − χn )u)2 (t)
((1 − χn )u(s) − (1 − χn )u(t))2
dt +
ds dt
t
(s − t)2
R+
R+ ×R+
)
( Z
(3.24)
¶2
µ
Z ∞ 2
ZZ
2n
1
u
(t)
u(s)
−
u(t)
ds dt +
dt .
≤C
u2 (t) dt +
2n 0
s−t
t
(n,∞)×R+
n
The last two terms above clearly tend to zero as n → ∞. To estimate the first term,
we integrate by parts (we may assume that u is smooth, it is the decay at infinity
that is the issue).
¶
Z 2n
Z 2n µZ 2n 2
Z t 2
1
1
u (s)
u (s)
2
u (t) dt =
ds −
ds dt
2n 0
2n 0
s
s
0
0
(3.25)
Z 2n Z 2n 2
Z
u (s)
log n 2n u2 (s)
1
ds dt +
ds,
≤
2n log n t
s
2n 0
s
which clearly tends to zero as n tends to infinity. The lemma follows.
1/2
We now give the following equivalent characterization of B0 (R+ ).
¤
594
Magnus Fontes
1/2
Lemma 3.8. A function u ∈ Lp (R+ ) belongs to B0 (R+ ) if and only if the
F00 (R+ )-distribution derivative
1/2
on B0 (R+ ) is given by
1/2
∂+ u
∂t1/2
∈ L2 (R+ ). Furthermore, an equivalent norm
° ∂ 1/2 u °
°
°
(3.26)
kuk = kukLp (R+ ) + ° +1/2 °
.
∂t
L2 (R+ )
Remark. We recall that the F00 (R+ )-distribution derivative, apart from what
happens inside R+ , also controls what happens on the boundary {0}. The fact that
1/2
∂+ u
∂t1/2
∈ L2 (R+ ) thus actually contains a lot of information about u’s behaviour at 0.
1/2
Proof. It is clear that a function in B0 (R+ ) has the F00 (R+ )-distribution de∂
1/2
u
rivative ∂t+1/2 in L2 (R+ ).
On the other hand, let E0 be the extension by zero operator. Then if u ∈ Lp (R+ )
and the F00 (R+ )-distribution derivative
Z
(3.27)
1/2
∂ Φ
E0 (u) − 1/2 dt =
∂t
R
1/2
∂+ u
∂t1/2
µ
Z
E0
R
∈ L2 (R+ ), we have
1/2 ¶
∂+ u
Φ dt ;
∂t1/2
∂
1/2
Φ ∈ F (R).
E (u)
This shows that the F 0 (R)-distribution derivative +∂t1/20
belongs to L2 (R). An
easy computation shows that
¯2
Z ¯¯ 1/2 ¯¯2
Z ¯¯ 1/2
¯
¯ ∂+ u ¯
¯ ∂+ E0 (u)(t) ¯
¯ dt
¯ 1/2 ¯ dt =
¯
¯
¯
∂t1/2
R+ ¯ ∂t
R¯
(3.28)
¯2
¯
ZZ
Z
¯ E0 (u)(s) − E0 (u)(t) ¯
E0 (u)2
¯
¯
∼
ds
dt
+
dt.
¯
¯
s−t
t
R+ ×R+
R+
Since E0 (u) = u on (0, ∞), the lemma follows.
¤
We now give a corresponding equivalent norm on B 1/2 (R+ ).
Lemma 3.9. If u ∈ B 1/2 (R+ ), then the F 0 (R+ )-distribution derivative
belongs to L2 (R+ ).
Furthermore, an equivalent norm on B 1/2 (R+ ) is given by
1/2
∂− u
∂t1/2
° ∂ 1/2 u °
°
°
.
kuk = kukLp (R) + ° −1/2 °
∂t
L2 (R+ )
Remark. In contrast to the F00 (R+ )-distribution derivative, the F 0 (R+ )-distribution derivative that we use in this definition “does not see” what happens on the
boundary, {0}.
(3.29)
Proof. Since F (R+ ) is dense in B 1/2 (R+ ), it is enough to show that
¯
¯
Z ¯¯ 1/2 ¯¯2
ZZ
¯ u(s) − u(t) ¯2
¯ ∂− u ¯
¯
¯
(3.30)
¯ 1/2 ¯ dt ∼
¯ s − t ¯ ds dt,
¯
R+ ¯ ∂t
R+ ×R+
for functions in F (R+ ), where ∼ means that the seminorms are equivalent.
Initial-boundary value problems for parabolic equations
595
For p = 2 we (temporarily) denote the closure of F (R+ ) in the norm
° ∂ 1/2 u °
°
°
(3.31)
kuk = ° −1/2 °
+ kukL2 (R+ ) ,
∂t
L2 (R+ )
by H.
It follows directely from the definitions, and the fact that F (R+ ) is dense in
1/2
B (R+ ), that B 1/2 (R+ ) is continuously imbedded in H.
We shall now show that in fact H = B 1/2 (R+ ).
1/2
+ u. Then T : B0 (R+ ) → H ∗ is continuLet T denote the operator T : u 7→ ∂u
∂t
ous. This follows from fractional integration by parts,
Ã
!
1/2
1/2
∂+ u ∂− Φ
(3.32)
hT u, Φi =
,
+ (u, Φ)L2 ; Φ ∈ F (R+ ), u ∈ F0 (R+ ),
∂t1/2 ∂t1/2
2
L
1/2
and the fact that F (R+ ) is dense in H and that F0 (R+ ) is dense in B0 (R+ ).
Now by the Hahn–Banach theorem, given ξ ∈ H ∗ there exist elements u, v ∈
L2 (R+ ) such that
Ã
!
1/2
∂ Φ
(3.33)
hξ, Φi = u, − 1/2
+ (v, Φ)L2 ; Φ ∈ F (R+ ).
∂t
2
L
We can thus extend ξ by zero to an element E0 (ξ) of B 1/2 (R)∗ . Since T : B 1/2 (R)
→ B 1/2 (R)∗ is an isomorphism, we can find a unique element u ∈ B 1/2 (R) such that
1/2
T u = E0 (ξ) in F 0 (R). But this holds if and only if u ∈ B0 (R+ ) and T u = ξ in
F00 (R+ ).
1/2
Thus T : B0 (R+ ) −→ H ∗ is an isomorphism.
Furthermore, by direct computation (or by interpolation (recall that p = 2)), we
know that
(3.34)
1/2
T : B0 (R+ ) −→ B 1/2 (R+ )∗
is an isomorphism.
Since F (R+ ) is densely continuously imbedded in both H and B 1/2 (R+ ) and
thus H ∗ and B 1/2 (R+ )∗ both are well defined subspaces in F00 (R+ ), we see that
H ∗ and B 1/2 (R+ )∗ are identical as subspaces of F00 (R+ ) and equivalent as Hilbert
spaces.
Since B 1/2 (R+ ) ,→ H, by Riesz representation theorem, this implies that H and
B 1/2 (R+ ) have equivalent norms.
From a scaling argument it now follows that
¯
¯
ZZ
Z ¯¯ 1/2 ¯¯2
¯ u(s) − u(t) ¯2
¯ ∂− u ¯
¯
¯
(3.35)
¯ 1/2 ¯ dt ∼
¯ s − t ¯ ds dt,
¯
¯
∂t
R+ ×R+
R+
for functions in B 1/2 (R+ ). The lemma follows.
4. Parabolic equations
We shall consider operators of the form
∂u
(4.1)
Tu =
− ∇x · A(x, t, ∇x u),
∂t
¤
596
Magnus Fontes
on a space-time cylinder Q+ = Ω × R+ , where Ω is an open and bounded set in Rn
with smooth boundary.
We shall assume the following structural conditions for the function A : Ω × R+ ×
n
R → Rn .
(1) Q+ 3 (x, t) 7→ A(x, t, ξ) is Lebesgue measurable for every fixed ξ ∈ Rn .
(2) Rn 3 ξ 7→ A(x, t, ξ) is continuous for almost every (x, t) ∈ Q+ .
(3) For every ξ, η ∈ Rn , ξ 6= η and almost every (x, t) ∈ Q+ , we have
(4.2)
(A(x, t, ξ) − A(x, t, η), ξ − η) > 0.
(4) There exists p ∈ (1, ∞), a constant λ > 0 and a function h ∈ L1 (Q+ ) such
that for every ξ ∈ Rn and almost every(x, t) ∈ Q+ :
(A(x, t, ξ), ξ) ≥ λ|ξ|p − h(x, t).
(4.3)
(5) There exists a constant Λ ≥ λ > 0 and a function H ∈ Lp/(p−1) (Q+ ) such
that for every ξ ∈ Rn and almost every (x, t) ∈ Q+ :
|A(x, t, ξ)| ≤ Λ|ξ|p−1 + H(x, t).
(4.4)
The Carathéodory conditions 1 and 2 above guarantee that the function Q 3 (x, t) 7→
A(x, t, Φ(x, t)) is measurable for every function Φ ∈ Lp (Q+ , Rn ). Condition 3 is a
strict monotonicity condition that gives us uniqueness results. Conditions 4 (coercivity) and 5 (boundedness) give us apriori estimates that imply existence results (see
[1]).
1/2
We now introduce some function spaces, and in their definitions ∂− /∂t1/2 should
be understood in the F 0 ·,· (Q) distribution-sense.
Definition 4.1. For 1 < p < ∞, set
(
)
1/2
∂
u
∂u
1,1/2
(4.5)
B·,·
(Q) = u ∈ Lp (Q); −1/2 ∈ L2 (Q),
∈ Lp (Q), i = 1, . . . , n .
∂t
∂xi
We equip these spaces with the following norms.
n °
° ∂ 1/2 u °
°
X
°
°
° ∂u °
(4.6)
kukB 1,1/2 (Q) = ° −1/2 °
+ kukLp (Q) +
°
° p .
·,·
∂t
∂x
L2 (Q)
L (Q)
i
i=1
Computing in F 0 ·,· (Q) we see that we can represent these spaces as closed subspaces of the direct sum L2 (Q) ⊕ Lp (Q) ⊕ · · · ⊕ Lp (Q), and thus they are reflexive
and separable Banach spaces in the topologies arising from the given norms.
Since the lateral boundary is smooth (in fact Lipschitz continuous suffices), we
1,1/2
can extend an element in B·,· (Q) to all of Rn × R and then cut off in the space
variables. By regularizing it is clear that functions smooth up to the boundary are
1,1/2
1,1/2
dense in B·,· (Q). To show that F·,· (Q) is dense in B·,· (Q) we only have to prove
that we can “cut off” in time. This will follow as in Lemma 3.4 once we have the
following result.
1,1/2
Lemma 4.1. If u ∈ B·,· (Q), then
¯
¯
ZZ
ZZZ
¯ u(x, s) − u(x, t) ¯2
¯
¯
(4.7)
¯ dx ds dt = 2π
¯
s−t
Ω×R×R
¯
¯
¯ ∂ 1/2 u ¯2
¯ − ¯
¯ 1/2 ¯ dx dt.
¯
Q ¯ ∂t
Initial-boundary value problems for parabolic equations
∂
1/2
597
u
Proof. That ∂t−1/2 = v means that
ZZ
ZZ
1/2
∂+ Φ(x, t)
(4.8)
u(x, t)
dx dt =
v(x, t)Φ(x, t) dx dt;
∂t1/2
Q
Q
Φ ∈ F0,· (Q).
Now for almost every x ∈ Ω, Ω 3 x 7→ u(x, ·) ∈ Lp (R) and Ω 3 x 7→ v(x, ·) ∈ L2 (R)
are well defined. Let S denote the set of common Lebesgue points. Since the Lebesgue
points of a function can only increase by multiplication with a smooth function, by
taking limits of mean values, we get that
Z
Z
1/2
∂+ Φ(x, t)
(4.9)
u(x, t)
dt =
v(x, t)Φ(x, t) dt; Φ ∈ F0,· (Q),
∂t1/2
R
R
for all x ∈ S. This implies that for almost every x ∈ Ω the Lp (R) function t 7→ u(x, t)
has half a derivative equal to v(x, t) ∈ L2 (R). So from the one-dimensional result it
follows that
¯
¯
ZZ
Z ¯¯ 1/2 ¯¯2
¯ u(x, s) − u(x, t) ¯2
¯ ∂− u ¯
¯
¯ ds dt = 2π
(4.10)
¯ 1/2 ¯ dt,
¯
¯
¯
s−t
R×R
R ¯ ∂t
for almost every x ∈ Ω. Integrating with respect to x, the lemma follows.
¤
We conclude that:
1,1/2
Lemma 4.2. The space of testfunctions F·,· (Q) is dense in B·,·
(Q).
We now introduce the following subspace that corresponds to zero boundary data
on the lateral boundary ∂Ω × R and as |t| → ∞.
1,1/2
1,1/2
Definition 4.2. Let B0,· (Q) denote the closure of F0,· (Q) in the B·,·
topology.
(Q)-
We shall work with the following two sets of function spaces on Q+ .
1,1/2
Definition 4.3. Let B∗,· (Q+ ) denote the space of functions defined on Q+
1,1/2
that can be extended to elements in B∗,· (Q).
1,1/2
Furthermore, let B∗,0 (Q+ ) denote the space of functions defined on Q+ that
1,1/2
can be extended by zero to elements in B∗,· (Q).
Here ∗ optionally stands for · or 0. A zero in the first position corresponds to zero
boundary data on the lateral boundary and a zero in the second position corresponds
to zero initial data.
1,1/2
1,1/2
Clearly B∗,0 (Q+ ) can be identified with a closed subspace of B∗,· (Q).
We give the following two simple lemmas concerning these spaces and, as in the
case of the real line, we omit the easy proofs.
1,1/2
Lemma 4.3. The function space B∗,0 (Q+ ) becomes a Banach space with the
norm
½Z
u2 (x, t)
p
p
dt dx
kukB 1,1/2 (Q+ ) = kukL (Q+ ) + k∇x ukL (Q+ ) +
∗,0
t
Q+
)1/2
(4.11)
µ
¶2
ZZZ
u(x, s) − u(x, t)
.
+
dx ds dt
s−t
Ω×R+ ×R+
598
Magnus Fontes
1,1/2
Lemma 4.4. The function space B∗,· (Q+ ) becomes a Banach space with the
norm
kukB 1,1/2 (Q+ ) = kukLp (Q+ ) + k∇x ukLp (Q+ )
·,·
(Z Z Z
)1/2
µ
¶2
(4.12)
u(x, s) − u(x, t)
+
dx ds dt
.
s−t
Ω×R+ ×R+
1,1/2
1,1/2
Furthermore, a continuous symmetric extension mapping from B∗,· (Q+ ) to B∗,· (Q)
is given by ES (u)(x, t) = u(x, |t|).
1,1/2
0
(Q+ ) we can give an equivalent characterization of B·,0 (Q+ ).
Computing in F·,0
1,1/2
Lemma 4.5. A function u ∈ Lp (Q+ ) belongs to B·,0 (Q+ ) if and only if
∂
1/2
u
0
the F·,0
(Q+ )-distribution derivative ∂t+1/2 belongs to L2 (Q+ ), and the F·,·0 (Q+ )-dis1,1/2
tribution derivatives ∇x u ∈ Lp (Q+ ). Furthermore, an equivalent norm on B·,0 (R+ )
is then given by
° ∂ 1/2 u °
°
°
(4.13)
kuk = k∇x ukLp (Q+ ) + kukLp (Q+ ) + ° +1/2 °
.
∂t
L2 (Q+ )
Proof. As on the real line.
¤
Using the corresponding result on the real half-line and the same type of argument
1,1/2
as in the proof of Lemma 4.1, we see that an equivalent norm on B∗,· (Q+ ) is given
by
n °
1/2
°
X
∂− u
° ∂u °
p
(4.14)
kuk = k 1/2 kL2 (Q+ ) + kukL (Q+ ) +
,
°
°
∂t
∂xi Lp (Q+ )
i=1
∂
1/2
where ∂t−1/2 is understood in the F 0 ·,· (Q+ )-distribution sense.
We have the following density results:
1,1/2
Theorem 4.1. The space of testfunctions F·,∗ (Q+ ) is dense in B·,∗ (Q+ ). Fur1,1/2
thermore, the space of testfunctions F0,∗ (Q+ ) is dense in B0,∗ (Q+ ).
Proof. Since the boundary of Ω is smooth we have good extension operators in
the space variables, and we can also translate the support of functions away from
the lateral boundary without spreading the support in the time direction. The result
thus follows exactly as in Lemma 3.7.
¤
We point out the following result that follows immediately from the given norms.
1,1/2
1,1/2
Lemma 4.6. The space B∗,0 (Q+ ) is continuously imbedded in B∗,· (Q+ ).
∂
1/2
u
∂
1/2
u
We also remark that the (semi)norms k −∂t kL2 (Q+ ) and k +∂t kL2 (Q+ ) are not
1,1/2
equivalent. In fact in Lemma 4.8 below we show that B0,0 (Q+ ) is a dense subspace
1,1/2
of B0,· (Q+ ). This is of course connected with the well known fact that if u ∈ L2 (Q)
∂
1/2
u
and ∂t−1/2 ∈ L2 (Q), it is in general impossible to define a trace on Ω×{0} (for instance
the function (x, t) 7→ log | log |t|| locally belongs to this space). Still a function in
1,1/2
B·,0 (Q+ ) is of course zero on Ω × {0} in the sense that
Initial-boundary value problems for parabolic equations
ZZ
(4.15)
Q+
599
u2 (x, t)
dxdt < ∞.
t
We shall now discuss homogeneous data on the whole parabolic boundary.
4.1. Homogeneous data. We introduce the following space of F 0 ·,· (Q)distributions defined globally in time, but supported in Q+ .
Definition 4.4. Let
(4.16)
−1,−1/2
(Q+ )
B·,0
n
:= ξ ∈
1,1/2
B0,· (Q)∗ ;
o
ξ = 0 in Ω × (−∞, 0) .
From Theorem 4.3 and Theorem 4.4 in [1] follows
Theorem 4.2. For T as defined in (4.1), satisfying the structural conditions
(1)–(5),
1,1/2
(4.17)
−1,−1/2
T : B0,0 (Q+ ) −→ B·,0
(Q+ )
is a bijection.
−1,−1/2
We shall now show that B·,0
1,1/2
B0,· (Q+ ).
(Q+ ) can be identified with the dual space of
−1,−1/2
Lemma 4.7. We can identify B·,0
1,1/2
(Q+ ) with B0,· (Q+ )∗ .
Remark. Note that we here identify a subspace of F 0 ·,· (Q) with a subspace of
F 0 ·,0 (Q+ ).
1,1/2
Proof. Given ξ ∈ B0,· (Q+ )∗ we have (by the Hahn–Banach theorem) u0 ∈
0
L2 (Q+ ) and ui ∈ Lp (Q+ ), i = 1, . . . , n, such that
ZZ
n
1/2
∂ Φ X ∂Φ
(4.18)
hξ, Φi =
u0 −
+
ui
dx dt; Φ ∈ F0,· (Q+ ).
∂t
∂x
i
Q+
i=1
It is thus clear that we can extend this ξ to all of F0,· (Q) by zero. Set
ZZ
n
1/2
∂− Φ X
∂Φ
(4.19)
hξ0 , Φi =
E0 (u0 )
+
E0 (ui )
dx dt; Φ ∈ F0,· (Q),
∂t
∂xi
Q
i=1
where E0 denotes the operator that extends a function with 0 to all of Q. The
1,1/2
−1,−1/2
mapping B0,· (Q+ )∗ 3 ξ 7→ ξ0 ∈ B·,0
(Q+ ) is clearly injective, but it is also
−1,−1/2
surjective. This follows since given ξ ∈ B·,0
(Q+ ), by Theorem 4.2 above, there
1,1/2
exists a (unique) uξ ∈ B0,0 (Q+ ) such that
∂uξ
− ∇x · (|∇x uξ |p−2 ∇x uξ ) = ξ,
∂t
(4.20)
i.e.,
ZZ
(4.21)
1/2
hξ, Φi =
Q
1/2
∂+ uξ ∂− Φ
+ (|∇x uξ |p−2 ∇x uξ ) · ∇x Φ dx dt;
∂t
∂t
and we see that ξ has the required form.
Thus we can reformulate Theorem 4.2.
Φ ∈ F0,· (Q),
¤
600
Magnus Fontes
Theorem 4.3. For T as defined in (4.1), satisfying the structural conditions
(1)–(5),
(4.22)
1,1/2
1,1/2
T : B0,0 (Q+ ) −→ B0,· (Q+ )∗
is a bijection.
1,1/2
Remark. This theorem of course means that given ξ ∈ B0,· (Q+ )∗ there exists
1,1/2
a unique u ∈ B0,0 (Q+ ) such that
(4.23)
hT (u), Φi = hξ, Φi ;
1,1/2
Φ ∈ B0,· (Q+ ).
Which means precisely that
ZZ
1/2
1/2
∂ + uξ ∂ − Φ
(4.24)
hξ, Φi =
+ A(x, t, ∇x u) · ∇x Φ dx dt;
∂t
∂t
Q+
Φ ∈ F0,· (Q+ ),
1,1/2
since F0,· (Q+ ) is dense in B0,· (Q+ ).
The following structure theorem for our source data space is an immediate consequence of the Hahn–Banach theorem.
1,1/2
Theorem 4.4. Given ξ ∈ B0,· (Q+ )∗ there exist functions u0 ∈ L2 (Q+ ) and
u1 , . . . , un ∈ Lp/(p−1) (Q+ ) such that
1/2
(4.25)
n
∂ u0 X ∂ui
ξ= +
+
∂t
∂xi
i=1
0
in F·,0
(Q+ ).
Our next result implies that in general it is actually enough to test our equations
with F0,0 (Q+ ) instead of F0,· (Q+ ).
Lemma 4.8. The continuous imbedding
(4.26)
1,1/2
1,1/2
B0,0 (Q+ ) ,→ B0,· (Q+ )
is dense.
1,1/2
Proof. It is enough to show that if ξ ∈ B0,· (Q+ )∗ and hξ, Φi = 0 for all
1,1/2
Φ ∈ B0,0 (Q+ ), then ξ = 0.
1,1/2
1,1/2
Now given ξ ∈ B0,· (Q+ )∗ , by Theorem 4.3, there exists a unique uξ ∈ B0,0 (Q+ )
such that
∂uξ
(4.27)
− ∇x · (|∇x uξ |p−2 ∇x uξ ) = ξ.
∂t
1,1/2
Now if hξ, Φi = 0 for all Φ ∈ B0,0 (Q+ ), then with Φ = uξ we get
ZZ
(4.28)
|∇x uξ |p dx dt = 0.
Q+
By the Poincaré inequality uξ = 0, and so ξ = 0.
¤
4.2. Non-homogeneous initial data. We will first introduce the space that
will carry the initial data. In the definition, all derivatives should be understood in
the F 0 ·,· (Q+ )-distribution sense.
Initial-boundary value problems for parabolic equations
601
Definition 4.5. Let
n
o
∂u
1,1/2
2
p0
−1,p0
(4.29) BI (Q+ ) = u ∈ B0,· (Q+ ) ∩ Cb ([0, ∞), L (Ω));
∈ L (R+ , W
(Ω)) .
∂t
Here Cb ([0, ∞), L2 (Ω)) denotes the space of bounded continuous functions from
0
0
∈ Lp (R+ , W −1,p (Ω)) means exactly that
[0, ∞) into L2 (Ω), and ∂u
∂t
∂Φ
i| ≤ Ck∇x ΦkLp (Q+ ) ; Φ ∈ F0,0 (Q+ ),
∂t
for some constant C > 0. The smallest possible constant is by definition k ∂u
k 0
∂t Lp (R+ ,
0
W −1,p (Ω)) .
We equip BI (Q+ ) with the following norm
° ∂u °
° °
(4.31)
kukBI (Q+ ) := kukB 1,1/2 (Q+ ) + sup ku(·, t)kL2 (Ω) + ° ° 0
.
0,·
∂t Lp (R+ ,W −1,p0 (Ω))
t∈R+
(4.30)
|hu,
Using Theorem 4.3 and the monotonicity of A(x, t, ·) we shall now prove that we
always have a unique solution in BI (Q+ ) to the following initial value problem.
Theorem 4.5. Given u0 ∈ L2 (Ω), there exists a unique element u ∈ BI (Q+ )
such that
∂u
(4.32a)
− ∇x · A(x, t, ∇x u) = 0 in F 0 ·,· (Q+ ),
∂t
(4.32b)
u = u0 on Ω × {0}.
Proof. Uniqueness follows immediately from the monotonicity of A(x, t, ·) by
pairing with a cut off function in time multiplied with the difference of two solutions.
To prove existence we first note that if u0 ∈ D(Ω), we can extend it for instance
to a smooth testfunction U0 ∈ D(Ω × (−2, 2)) such that U0 (x, t) = u0 (x) when
−1 < t < 1.
1,1/2
0
Since ∂U
∈ B0,· (Q+ )∗ , by Theorem 4.3, we know that there exists a unique
∂t
1,1/2
w ∈ B0,0 (Q+ ) such that
(4.33)
∂w
∂U0
− ∇x · A(x, t, ∇x w + ∇x U0 ) = −
∂t
∂t
1,1/2
in B0,· (Q+ )∗ .
1,1/2
Then clearly u = (w + U0 ) ∈ B0,· (Q+ ) solves (4.32), and the initial value is taken
in the sense that
ZZ
(u(x, t) − u0 (x))2
(4.34)
dx dt < ∞.
t
Ω×(0,1)
By standard arguments it follows from (4.32) that u ∈ BI (Q+ ) and so the initial
data is actually taken in Cb ([0, ∞), L2 (Ω))-sense.
Given u0 ∈ L2 (Ω) we now choose a sequence D(Ω) 3 un0 → u0 in L2 (Ω).
Let un denote the solution of (4.32) with initial data un0 . By testing with un χ,
where χ is a standard cut off function in time, in (4.32), we get that
Z
Z
n
m 2
2
(4.35)
sup (u − u ) (x, t) dx ≤ (un0 − um
0 ) (x) dx.
t∈R+
Ω
Ω
n
It is also clear that k∇x u kLp (Q+ ) is bounded by a constant independent of n.
Finally we note that we can extend un symmetrically to Q and the extended
n
0
0
1,1/2
function ES (un ) ∈ B0,· (Q) will satisfy ∂ES∂t(u ) ∈ Lp (R, W −1,p (Ω)).
602
Magnus Fontes
We then have
Z Z 1/2
Z D
1/2
E
∂− ES (un ) ∂− Φk
∂ES (un )
(4.36)
dx
dt
=
,
h(Φ
)
dt,
k
∂t1/2
∂t1/2
∂t
Q
R
1,1/2
for a sequence F0,· (Q) 3 Φk → ES (un ) in B0,· (Q).
∂
1/2
E (un )
S
This implies that k − ∂t1/2
kL2 (Q+ ) is bounded by a constant independent of n.
We conclude that kun kBI (Q+ ) ≤ C, where C < ∞ is a constant independent of n.
We can now extract a weakly convergent subsequence and in fact, as we have
seen, we actually have strong convergence in Cb ([0, ∞), L2 (Ω)) and thus the limit
function satisfies the initial conditions.
Finally a Minty argument using the monotonicity of A(x, t, ·) shows that the limit
function solves (4.32). The theorem follows.
¤
4.3. Fully non-homogeneous initial-boundary values. We shall now introduce the function space that will carry both initial and lateral boundary data.
1,1/2
1,1/2
Since we have continuous imbeddings B0,· (Q+ ) ,→ B·,· (Q+ ) and BI (Q+ ) ,→
1,1/2
B·,· (Q+ ), the following definition makes sense.
Definition 4.6. Let
1,1/2
X 1,1/2 (Q+ ) = B·,0 (Q+ ) + BI (Q+ )
(4.37)
be equipped with the norm
(4.38)
³
kukX 1,1/2 (Q+ ) =
inf
(u1 ,u2 )∈Ku
´
ku1 kB 1,1/2 (Q+ ) + ku2 kBI (Q+ ) ,
·,0
where the infimum is taken over the set
o
n
1,1/2
(4.39)
Ku = (u1 , u2 ); u1 + u2 = u, u1 ∈ B·,0 (Q+ ), u2 ∈ BI (Q+ ) .
The following imbeddings are immediate
(4.40)
kukX 1,1/2 (Q+ ) ≤ kukB 1,1/2 (Q+ ) ;
(4.41)
kukX 1,1/2 (Q+ ) ≤ kukBI (Q+ ) ;
(4.42)
kukB 1,1/2 (Q+ ) ≤ CkukX 1,1/2 (Q+ ) ;
·,0
1,1/2
u ∈ B·,0 (Q+ ),
u ∈ BI (Q+ ),
·,·
u ∈ X 1,1/2 (Q+ ).
For an element in X 1,1/2 (Q+ ) we can always define the trace on Ω × {0}.
Theorem 4.6. There exists a continuous linear and surjective trace operator
(4.43)
T r0 : X 1,1/2 (Q+ ) −→ L2 (Ω).
There also exists a bounded extension operator
(4.44)
E : L2 (Ω) −→ X 1,1/2 (Q+ )
such that T r0 ◦ E = IdL2 (Ω) .
1,1/2
Proof. Given u ∈ X 1,1/2 (Q+ ), there exist u1 ∈ B·,0 (Q+ ) and u2 ∈ BI (Q+ )
such that u = u1 + u2 . Since u2 ∈ BI (Q+ ) =⇒ u2 ∈ Cb ([0, +∞), L2 (Ω)), u2 |Ω×{0}
is a well defined element of L2 (Ω). We now define u|Ω×{0} = u2 |Ω×{0} . We have to
show that this is independent of the decomposition of u, but if we have two different
Initial-boundary value problems for parabolic equations
603
1,1/2
decompositions u1 + u2 = v1 + v2 as above, then (u2 − v2 ) ∈ BI (Q+ ) ∩ B·,0 (Q+ ),
which implies that
ZZ
(u2 − v2 )2 (x, t)
dx dt < +∞,
(4.45)
t
Ω×(0,+∞)
and so u2 (·, 0) = v2 (·, 0) since they both belong to Cb ([0, +∞), L2 (Ω)).
Now
(4.46)
ku(·, 0)kL2 (Ω) = ku2 (·, 0)kL2 (Ω) ≤ Cku2 kBI (Q+ ) ,
for any decomposition u = u1 + u2 as above. Taking the infimum over all such
decompositions gives:
(4.47)
ku(·, 0)kL2 (Ω) ≤ CkukX 1,1/2 (Q+ ) ,
u ∈ X 1,1/2 (Q+ ).
Now given u0 ∈ L2 (Ω), let E(u0 ) be the (unique) solution in BI (Q+ ) of the initial
value problem:
∂u
(4.48a)
− ∇x · (|∇x u|p−2 ∇x u) = 0 in Q+ = Ω × R+ ,
∂t
(4.48b)
u = u0 on Ω × {0}.
Clearly this extension map satisfies T r0 ◦ E = IdL2 (Ω) and furthermore
(4.49)
kE(u0 )kBI (Q+ ) ≤ Cku0 kL2 (Ω) ,
and thus
(4.50)
kE(u0 )kX 1,1/2 (Q+ ) ≤ Cku0 kL2 (Ω) .
¤
Remark. Note that if p = 2, the extension map is linear.
Theorem 4.7. We have the following imbedding:
kukB 1,1/2 (Q+ ) ≤ CkukX 1,1/2 (Q+ ) ;
(4.51)
·,0
1,1/2
u ∈ B·,0 (Q+ ).
1,1/2
B·,0 (Q+ )
1,1/2
Proof. If u ∈
and u = u1 + u2 with u1 ∈ B·,0 (Q+ ) and u2 ∈ BI (Q+ ),
1,1/2
then u2 (·, 0) = 0 since u2 ∈ B·,0 (Q+ ) ∩ BI (Q+ ). Thus u2 can be extended by zero
to all of Q. Since, by a continuity argument,
Z D
° ∂ 1/2 u °2
E
∂u2
° + 2°
1,1/2
(4.52)
=−
, h(u2 ) dt, u2 ∈ B·,0 (Q+ ) ∩ BI (Q+ ).
°
° 2
∂t
∂t
L (Q+ )
R+
We get
(4.53)
³
´
ku1 kB 1,1/2 (Q+ ) + ku2 kBI (Q+ ) ≥ C ku1 kB 1,1/2 (Q+ ) + ku2 kB 1,1/2 (Q+ )
·,0
·,0
·,0
≥ Cku1 + u2 kB 1,1/2 (Q+ ) = CkukB 1,1/2 (Q+ ) ,
·,0
·,0
where C > 0. Taking the infimum concludes the proof.
¤
We immediately get the following
Corollary 4.1. There exist constants C1 , C2 > 0 such that
(4.54)
C1 kukB 1,1/2 (Q+ ) ≤ kukX 1,1/2 (Q+ ) ≤ C2 kukB 1,1/2 (Q+ ) ;
0,0
1,1/2
0,0
u ∈ F0,0 (Q+ ).
Thus B0,0 (Q+ ) is the closure of F0,0 (Q+ ) in the X 1,1/2 (Q+ )-norm topology.
We are now ready to state our main theorem.
604
Magnus Fontes
1,1/2
Theorem 4.8. Given f ∈ B0,· (Q+ )∗ and g ∈ X 1,1/2 (Q+ ), there exists a unique
element u ∈ X 1,1/2 (Q+ ) such that
(4.55a)
(4.55b)
∂u
− ∇x · (A(x, t, ∇x u)) = f in F·,·0 (Q+ ),
∂t
1,1/2
u − g ∈ B0,0 (Q+ ).
Proof. Let w = u − g. Then (4.55) is equivalent to
(4.56a)
(4.56b)
∂w
∂g
− ∇x · (A(x, t, ∇x (w + g))) = f −
in F·,·0 (Q+ ),
∂t
∂t
1,1/2
w ∈ B0,0 (Q+ ).
1,1/2
Here ∂g
∈ F 0 ·,· (Q+ ) has a unique extension to an element in B0,· (Q+ )∗ . In fact, if
∂t
1,1/2
g ∈ X 1,1/2 (Q+ ), we can write g = g1 + g2 , where g1 ∈ B·,0 (Q+ ) and g2 ∈ BI (Q+ ).
Thus
∂Φ
∂Φ
∂Φ
|hg,
i| = |hg1 ,
i + hg2 ,
i|
∂t
∂t
(4.57)
´
³ ∂t
≤ C kg1 kB 1,1/2 (Q+ ) + kg2 kBI (Q+ ) kΦkB 1,1/2 (Q+ ) ;
·,0
0,·
Φ ∈ F0,0 (Q+ ).
1,1/2
Since, by Lemma 4.8 and Theorem 4.1, F0,0 (Q+ ) is dense in B0,· (Q+ ), it is clear
that we have a unique extension. If the function A(·, ·, ·) satisfies the structural
conditions 1–5 given above, then also A(·, ·, · + g), with g ∈ X 1,1/2 (Q+ ), satisfies the
same structural conditions (with new constants λ, Λ and functions H, h depending on
g). Thus Theorem 4.3, and the remark following Theorem 4.3, tell us that (4.56) has
a unique solution. This implies that u = w + g is the unique solution to (4.55). ¤
Remark. Note that since D(Q+ ) is densely continuously imbedded in F0,0 (Q+ )
it is equivalent to demand that (4.55) should hold in D 0 (Q+ ).
We shall conclude with a comment on the linear case.
The function spaces we have introduced so far coincides with well known function
spaces existing in the literature when p = 2. When p = 2 we shall follow existing
notation and replace B with H for all spaces (for instance, if p = 2 we shall write
1,1/2
1,1/2
H0,· (Q+ ) instead of B0,· (Q+ ) and so on).
1/2,1/4
The Sobolev space H·,·
(∂Ω × R+ ) below is defined by pull-backs in local
charts on ∂Ω.
Theorem 4.9. If p = 2 there exists a linear, continuous and surjective trace
operator
(4.58)
1/2,1/4
(∂Ω × R+ ).
T r : X 1,1/2 (Q+ ) −→ H·,·
There also exists a continuous and linear extension operator
(4.59)
1/2,1/4
(∂Ω × R+ ) −→ X 1,1/2 (Q+ ),
E : H·,·
such that T r ◦ E = Id|H 1/2,1/4 (∂Ω×R) .
·,·
Proof. Using a partition of unity argument and the Fourier multiplier operators
(4.60)
ms (D)u = ((1 + i2πτ + 4π 2 |ξ|2 )−s û)∨ ;
s ∈ R,
Initial-boundary value problems for parabolic equations
605
which preserves forward support in time, and have the property that
¡
¢
2s,s
(4.61)
ms (D) L2 (Rn × R) = H·,·
(Rn × R),
we can construct continuous linear operators:
1,1/2
1/2,1/4
T r : H·,0 (Q+ ) −→ H·,·
(∂Ω × R+ )
(4.62)
and
1,1/2
1/2,1/4
E : H·,·
(∂Ω × R+ ) −→ H·,0 (Q+ ),
(4.63)
such that T r ◦E = Id|H 1/2,1/4 (∂Ω×R) . Now given u ∈ X 1,1/2 (Q+ ), let u = u1 +u2 where
·,·
1,1/2
u1 ∈ H·,0 (Q+ ) and u2 ∈ HI (Q+ ). We define u|∂Ω×R+ = u1 |∂Ω×R+ . This definition
is independent of the decomposition of u. In fact, if u1 + u2 = v1 + v2 are two
decompositions as above, then u1 − v1 ∈ L2 (R+ , H01 (Ω)), and so (u1 − v1 )|∂Ω×R = 0.
Now
(4.64)
kT r(u)kH 1/2,1/4 (∂Ω×R+ ) ≤ Cku1 kH 1,1/2 (Q+ )
·,·
·,0
for any decomposition. Taking the infimum proves the continuity of T r. The
1,1/2
continuity of the extension operator E follows from the imbedding H·,0 (Q+ ) ,→
X 1,1/2 (Q+ ).
¤
Combining our trace theorems with Theorem 4.8 gives us in the linear case:
Theorem 4.10. If
∂u
− ∇x · (A(x, t, ∇x u)),
∂t
is a linear operator, satisfying the structural conditions 1–5 above, then
(4.65)
(4.66)
Tu =
X 1,1/2 (Q+ ) 3 u 7→ (T u, u|∂Ω×R+ , u|Ω×{0} )
1,1/2
1/2,1/4
∈ H0,· (Q+ )∗ × H·,·
(∂Ω × R+ ) × L2 (Ω)
is a linear isomorphism.
Acknowledgements. I thank Johan Råde for useful remarks and stimulating discussions in connection with this work, and Anders Holst, Per-Anders Ivert and Stefan
Jakobsson for reading and commenting on this paper.
References
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Differential Equations 25:3&4, 2000, 681–702.
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Report no. 22, 2000.
[3] Kaplan, S.: Abstract boundary value problems for linear parabolic equations. - Ann. Sc. Norm.
Super. Pisa Cl. Sci. (4) 20, 1966, 395–419.
[4] Ladyzenskaja, O. A., V. A. Solonnikov, and N. N. Uralceva: Linear and quasi-linear
equations of parabolic type. - Transl. Math. Monogr. 23, AMS, Providence, Rhode Island, 1968.
[5] Lions, J. L., and E. Magenes: Problémes aux limites non homogénes et applications I–II. Dunod, Paris, 1968.
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Received 26 November 2008